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Mar 8, 2012 at 18:36 comment added Ulrich Pennig Okay, but still. If you have $1 \in B$, then $1 = p \cdot 1 \cdot p = p^2 = p$, so $B$ can not be unital. Or am I misinterpreting something?
Mar 8, 2012 at 14:38 comment added Aviv Oh, Iam sorry, I mean that the projection $p$ is not trivial.
Mar 8, 2012 at 12:41 history answered Ulrich Pennig CC BY-SA 3.0